A string along which waves can travel is 2.70 m long and has a mass of 130 g. The tension in the string is 36.0 N What must be the frequency of traveling waves of amplitude 7.70 mm for the average power to be 170 W?

Answer

To solve the problem step by step, we will follow the outlined approach to find the frequency of the traveling waves on the string.

### Step 1: Calculate the mass per unit length (μ) of the string
The mass per unit length (μ) is calculated using the formula:
\[ \mu = \frac{m}{L} \]
where \( m \) is the mass of the string and \( L \) is its length.

Given:
- Mass \( m = 130 \, \text{g} = 0.130 \, \text{kg} \) (conversion from grams to kilograms)
- Length \( L = 2.70 \, \text{m} \)

Calculating \( \mu \):
\[ \mu = \frac{0.130 \, \text{kg}}{2.70 \, \text{m}} = 0.04815 \, \text{kg/m} \]

### Step 2: Calculate the wave speed (v) on the string
The wave speed (v) is given by the formula:
\[ v = \sqrt{\frac{T}{\mu}} \]
where \( T \) is the tension in the string.

Given:
- Tension \( T = 36.0 \, \text{N} \)

Calculating \( v \):
\[ v = \sqrt{\frac{36.0 \, \text{N}}{0.04815 \, \text{kg/m}}} \approx \sqrt{747.92} \approx 27.4 \, \text{m/s} \]

### Step 3: Use the average power formula to find angular frequency (ω)
The average power (P) of a wave can be expressed as:
\[ P = \frac{1}{2} \mu A^2 \omega^2 v \]
where:
- \( A \) is the amplitude of the wave.

Given:
- Average power \( P = 170 \, \text{W} \)
- Amplitude \( A = 7.70 \, \text{mm} = 0.0077 \, \text{m} \) (conversion from mm to m)

Rearranging the formula to solve for \( \omega^2 \):
\[ \omega^2 = \frac{2P}{\mu A^2 v} \]

Substituting the known values:
\[ \omega^2 = \frac{2 \times 170 \, \text{W}}{0.04815 \, \text{kg/m} \times (0.0077 \, \text{m})^2 \times 27.4 \, \text{m/s}} \]

Calculating \( \omega^2 \):
\[ \omega^2 = \frac{340}{0.04815 \times 0.00005929 \times 27.4} \approx \frac{340}{0.000079} \approx 4300000 \]
\[ \omega \approx \sqrt{4300000} \approx 2073.64 \, \text{rad/s} \]

### Step 4: Convert angular frequency (ω) to frequency (f)
The relationship between angular frequency (ω) and frequency (f) is given by:
\[ \omega = 2 \pi f \]
Rearranging to find \( f \):
\[ f = \frac{\omega}{2 \pi} \]

Substituting the value of \( \omega \):
\[ f = \frac{2073.64}{2 \pi} \approx \frac{2073.64}{6.2832} \approx 330.0 \, \text{Hz} \]

### Final Answer
The frequency of traveling waves must be approximately **330 Hz**.

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